Chapter 11.3, Problem 6CYP

To determine

Find $a_1$ in a geometric series for which ${\text{S}}_{\text{n}}=-26,240;n=8;r=-3$

Answer to Problem 6CYP

  $a_1=16$

Explanation of Solution

Given:

Find $a_1$ in a geometric series for which ${\text{S}}_{\text{n}}=-26,240;n=8;r=-3$

Concept Used:

The general summation formula for GP series is ${\text{S}}_{\text{n}}=\frac{a_1(r^n-1)}{r-1}$

Calculation:

  ${\text{S}}_{\text{n}}=-26,240;n=8;r=-3;\text{Find}{a}_1$

  $\begin{array}{l}{\text{S}}_{\text{n}}=-26,240;n=8;r=-3;\text{Find}{a}_1\\ {\text{S}}_{\text{n}}=\frac{a_1(r^n-1)}{r-1}\\ -\text{26,240}=\frac{a_1([{(-3)}^8-1]}{-3-1}=\frac{a_1·6560}{-4}=-1640·a_1\\ \frac{-1640·a_1}{-1640}=\frac{-26240}{-1640}\\ a_1=16\end{array}$

Thus, $a_1=16$